Csdrhrt

In a text book I am following, there is a definition of positive definite matrices, which I did not see before:

$x^TAx geq alpha x^Tx$

where $alpha$ is a positive scalar and $x in mathbb{R^n}$.

The usual definition I know is for every non-zero vector $x in mathbb{R^n}$, it is $x^TAx > 0$.

These definitions must be equivalent. It is trivial to show that the first definition implies $x^TAx > 0$. But I failed to see a way to show the other way around of the equivalence. I thought of applying SVD to the matrix $A$ and tried to find a way to show that $x^TAx$ is always lower bounded by a $alpha x^Tx$ with $alpha$ being positive, using singular values and vectors of $A$ but couldn't move forward. What is the correct way of approach here?

(The other question in the board assumes symmetric matrices. This one does not).

Let $f(x)=x^TAx$. Furthermore let $f(x)>0$ for all non-zero $x$.

Let $C:={x in mathbb R^n: ||x||_2=1}$. Then $C$ is compact and $f$ is continuous on $C$. Thus there is $x_0 in C$ such that

$f(x_0) le f(x)$ for all $x in C$.

Now put $ alpha =f(x_0)$. Then $alpha >0$.

It is your turn to show that $x^TAx geq alpha x^Tx$ for all $x in mathbb R^n$.

Hint: for $t in mathbb R^n$ and $t in mathbb R$ we have $f(tx)=t^2f(x).$

The following is valid only if the matrix $A$ doesn't have complex values, which is at least true if it is also symmetric.

Since $Asucc 0$, we can use it's diagonalisation as $A=USigma U^{dagger}$ such that $U U^dagger = U^dagger U = I$ and $Sigma$ is the diagonal matrix containing the eigenvalues of $A$, then for any $a>0$, $A-aI = U(Sigma - aI) U^dagger$ and so the eigenvalues of $A-aI$ are those of $A$ minus $a$.

Being positive definite means that the eigenvalues are strictly positive and so we can find $0<aleqmin_{i} lambda_i$ where $lbracelambda_irbrace_{iin[1{:}n]}$ are the eigen values such that the eigen values of $A-aI$ are positive which is $Asucc aI$ and hence $x^dagger A x geq a x^dagger x$